If 34.2 grams of lithium react with excess water, how many liters of hydrogen gas can be produced at 299 kelvin and 1.21 atmospheres? show all of the work used to solve this problem.

2li (s) + 2h2o (l) yields 2lioh (aq) + h2 (g)

Convert the grams of Li to moles: (34.2g)/(6.941 g/mol) = 4.927 moles of Li Looking at the balanced eqn:2 moles of Li react to give 1 mol of H2.  Using the ratio 2:1,4.927 mol of Li will react to give 2.464 mol of H2. You have to use the ideal gas law now: PV=nRT to solve for the volume the 2.464 mol of H2 will occupy: PV=nRT(1.21)(v) = (2.464)(0.0821)(299) Solving for V algebraically, V=50.0L

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Is there any options, for this